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Official Solution:If \(x\) is a positive integer, is \(y^2 (x^3  x + 1) \gt 75?\) We need to know values of two variables to answer the question:  The value of \(x\)
 The value of \(y\)
Statement (1) by itself is insufficient. We only know that \(y \gt 1\). Statement (2) by itself is insufficient. We know the value of \(x\). Knowing that \(x\) is a positive integer, we conclude that \(x1 = 4\), thus, \(x = 5\). However, we don't know the value of \(y\). Statements (1) and (2) combined are sufficient. If \(x = 5\), then \(5^3  5 + 1 = 121\). If \(y \gt 1\), then the statement will always be greater than 75. Answer: C
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Re: M0221
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11 Oct 2014, 03:55
from (2): we can also get the x=3. Bunuel can i infer this?



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11 Oct 2014, 04:00



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Bunuel wrote: Official Solution: We need to know values of two variables to answer the question:  The value of \(x\)
 The value of \(y\)
Statement (1) by itself is insufficient. We only know that \(y \gt 1\). Statement (2) by itself is insufficient. We know the value of \(x\). Knowing that \(x\) is a positive integer, we conclude that \(x1 = 4\), thus, \(x = 5\). However, we don't know the value of \(y\). Statements (1) and (2) combined are sufficient. If \(x = 5\), then \(x = 5^3  5 + 1 = 121\). If \(y \gt 1\), then the statement will always be greater than 75. Answer: C I'm not sure I fully understand this. If \(x = 5\) then, then \(x = 5^3  5 + 1 = 121\), as such, \(y^2 > 75/121\) which is always true, whether \(y = 1\) or \(y = 1\). That should be the second statement of x=5 is sufficient.



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Re: M0221
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11 Apr 2015, 08:15
gmatser1 wrote: Bunuel wrote: Official Solution: We need to know values of two variables to answer the question:  The value of \(x\)
 The value of \(y\)
Statement (1) by itself is insufficient. We only know that \(y \gt 1\). Statement (2) by itself is insufficient. We know the value of \(x\). Knowing that \(x\) is a positive integer, we conclude that \(x1 = 4\), thus, \(x = 5\). However, we don't know the value of \(y\). Statements (1) and (2) combined are sufficient. If \(x = 5\), then \(x = 5^3  5 + 1 = 121\). If \(y \gt 1\), then the statement will always be greater than 75. Answer: C I'm not sure I fully understand this. If \(x = 5\) then, then \(x = 5^3  5 + 1 = 121\), as such, \(y^2 > 75/121\) which is always true, whether \(y = 1\) or \(y = 1\). That should be the second statement of x=5 is sufficient. What if y=0? Or y=1/2? Or y=1/2? There are infinitely many values of y, for which y^2>75/121 is not true.
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Re: M0221
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14 Apr 2015, 23:50
Yes even I had the same question , but if we read the question carefully it says x is a positive integer. So, x=3 cannot be considered in this case.



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27 Feb 2016, 08:21
I think this is a highquality question and I don't agree with the explanation. Hi,
Since y is squared in the statement, a negative integer will not matter, right? Hence, shouldn't statement 2 alone be sufficient?



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29 Feb 2016, 09:10



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22 Aug 2016, 14:09
I think this is a highquality question and I agree with explanation.



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01 Dec 2016, 09:15
I think this is a highquality question and I don't agree with the explanation. As per Statement 2  X may have 2 possible solutions 5 and 3 so even if we analyse both statement together than also the whole value might not be greater than 75



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01 Dec 2016, 09:43



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Hi Bunuel, For this question, since the question asks whether or not y>0 and through these two statements we get (y^2)>75/121 which means that the solution set for y is (infinity to (75/121)^(1/2)) U ((75/121)^(1/2) to infinity). Am I missing something here?



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I solved statement 2 in following manner:
(x1)^2=16 x^22x+1=16 x^22x=15 x(x2)=15 So x=0 or x=17
Where did I go wrong?



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20 Jun 2017, 01:19
disharupani wrote: I solved statement 2 in following manner:
(x1)^2=16 x^22x+1=16 x^22x=15 x(x2)=15 So x=0 or x=17
Where did I go wrong? You can equate multiples to 0, when the whole product equals 0. For example, if it were x(x  2) = 0, then yes, x = 0 or x = 2. Notice that if x = 0, then x(x  2) = 15, does not hold. Also, it's not clear how you got x = 17, there. Again, if x = 17, then x(x  2) = 15, does not hold.
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Re: M0221
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29 Aug 2017, 12:39
Hi Bunuel,
1 clarification needed.. in statement 2> we get x=5 y^2 * 121>75 to be proven? y^2 > or = to zero even if we take y as 1/2 then y^2 =1/4 1/4 *121=30.25 <75 we get answer as NO
Put y=0 then y^2=0 0*121=0 <75 we get answer as NO
Put y=12/11 then y^2=144/121 144/121 *121=144 >75 we get answer as YES
if y= integer then definitely y^2 * 121>75
Please let me know what am I lacking in understanding?Acc. to me statement2 is sufficient.
Thanks, Stuti



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29 Aug 2017, 21:45
stutimittal04 wrote: Hi Bunuel,
1 clarification needed.. in statement 2> we get x=5 y^2 * 121>75 to be proven? y^2 > or = to zero even if we take y as 1/2 then y^2 =1/4 1/4 *121=30.25 <75 we get answer as NO
Put y=0 then y^2=0 0*121=0 <75 we get answer as NO
Put y=12/11 then y^2=144/121 144/121 *121=144 >75 we get answer as YES
if y= integer then definitely y^2 * 121>75
Please let me know what am I lacking in understanding?Acc. to me statement2 is sufficient.
Thanks, Stuti For some values of y, the answer is YES, for example if y = 100. For some values of y, the answer is NO, for example if y = 0. Two answers, hence insufficient.
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31 Aug 2017, 09:58
Ohok!! Thanks Sent from my Redmi Note 3 using GMAT Club Forum mobile app



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25 Oct 2017, 00:08
What do you guys think about the variable approach from Math Revolution? I think that in some cases it is better to use this approach, but in most cases  not. And if you are aiming for, at least, Q50, variable approach should not be your friend







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